Question

A ball is thrown vertically upward. It has a speed of
10 m/s when it has reached one half of its maximum
height. How high does the ball rise? (Taking
g= 10 m/s2)
[AIPMT (Prelims)-2005]
(1) 15 m
(2) 10 m
(3) 20 m
(4) 5 m​

Answer 1

Answer:-

h = 10 m

Option - 2

Given :-

v = 0m/s

u = 10 m/s

g = 10 m/s²

To find:-

The maximum height it reach.

Solution:-

when the ball reaches to one half of its maximum height the speed is 10 m/s.

Acceleration due to gravity is negative because of upward motion.

a = -g

We have equation of motion :-

★ $\boxed{\sf{ 2gh = v^2 - u^2}}$

$2 \times g \times \dfrac{h}{2}=v^2 - u^2$

put the given value,

→$-10 \times h = 0^2 + 10^2$

→$-10 \times h = -100$

→$h = \dfrac{-100}{-10}$

→$h = 10 m$

hence,

The ball rise up to 10 m.

Answer 2

Answer:

Maximum height that the ball will reach = 10 m

Explanation:

Given that,

a ball is thrown vertically upward

its speed = 10 m/s

when it covered the half maximum height

so,

height = h/2

now,

we have,

initial velocity(u) = 10 m/s

gravitational acceleration(g) = -10 m/s²

final velocity(v) = 0 m/s

.so,

by the gravitational equation of motion

v² = u² + 2gh

here,. height = h/2

putting the values,

(0)² = (10)² + 2(-10)h/2

-10h + 100 = 0

10h = -100

h = -100/-10

h = 10 m

so,

Maximum height that the ball will reach

= 10 m